Total number of orthonomal list for a linearly independent set of $m$ vectors

Question

Given a linearly independent list of $m$ vectors (none of them are orthogonal to each other), if we apply Gram-Schmidt orthogonalization to each of the $m!$ orderings then will each time I will get a different orthonomal list i.e total $m!$ orthonormal lists, or some of them will be same ?

Answer 1

They will indeed all be different. For suppose that two orderings $(v_{i_1},v_{i_2}, \ldots ,v_{i_m})$ and $(v_{j_1},v_{j_2}, \ldots ,v_{j_m})$ give the same basis.

Then the first vector is $\frac{v_{i_1}}{||v_{i_1}||}=\frac{v_{j_1}}{||v_{j_1}||}$. If $i_1\neq j_1$, the two vectors $v_{i_1}$ and $v_{j_1}$ would be independent, and this equality would be impossible. So we mut have $i_1=j_1$.

By induction, one sees then that $i_k=j_k$ for all $k$.